Four charges,each of magnitude q coulomb,are | vedclass

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(A) The four charges are located at $A(1,0,0), B(-1,0,0), C(0,1,0),$ and $D(0,-1,0)$. We need to find the electric field at point $P(0,0,1)$.The dista

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$\frac{1}{2 \sqrt{2}} \frac{q}{\pi \varepsilon_0} 	ext{ N/C}$

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(A) The four charges are located at $A(1,0,0), B(-1,0,0), C(0,1,0),$ and $D(0,-1,0)$. We need to find the electric field at point $P(0,0,1)$.The distance $r$ of point $P$ from each charge is $r = \sqrt{1^2 + 1^2} = \sqrt{2} 	ext{ m}$.The magnitude of the electric field due to each charge at point $P$ is:$E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{2}$.Due to symmetry,the horizontal components of the electric fields from the four charges cancel each other out. Only the vertical components (along the $Z$-axis) contribute to the net electric field.The angle $	heta$ that the electric field vector from each charge makes with the $Z$-axis is given by $\cos 	heta = \frac{1}{r} = \frac{1}{\sqrt{2}}$.The vertical component of the electric field from each charge is $E_z = E \cos 	heta = \left( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{2} ight) \cdot \frac{1}{\sqrt{2}}$.Since there are four such charges,the net electric field $E_{	ext{net}}$ is:$E_{	ext{net}} = 4 \cdot E_z = 4 \cdot \left( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{2 \sqrt{2}} ight) = \frac{q}{2 \sqrt{2} \pi \varepsilon_0} 	ext{ N/C}$.

Four charges,each of magnitude $q$ coulomb,are placed at points $(-1,0,0), (1,0,0), (0,-1,0)$ and $(0,1,0)$ in the $xy$-plane. The distances along the axes are measured in metres. The magnitude of the electric field at the point $(0,0,1)$ on the $Z$-axis is

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